Question

Derive an expression for the axial magnetic field of a finite solenoid of length $2l$ and radius $r$ carrying $l$. Under what condition does the field becomes equivalent to that produced by a bar magnet?

Answer 1

Consider a solenoid of length $2I$, radius $r$ and carrying a current $I$ and having $n$ turns per unit length.

Consider a point $P$ at a distance a from the center $O$ of solenoid.

Consider an element of solenoid of length $dx$ at a distance $x$ from its center.

This element is a circular current loop having $\left(ndx\right)$ turns. The magnetic field at axial point $P$ due to this current loop is

$dB=\frac{{\mu }_0\left(ndx\right)I{r}^2}{2(r^2+(a-x{)}^2{)}^{3/2}}$

The total magnetic field due to entire solenoid is

$\therefore B={\int }_{-1}^{+1}\frac{{\mu }_0\left(ndx\right)I{r}^2}{2(r^2+(a-x{)}^2{)}^{3/2}}$

For $a>l$ and $a>>r$, we have ${\{r^2+(a-x{)}^2\}}^{3/2}=a^3$

$\therefore B=\frac{{\mu }_{0}nL{r}^2}{2a^3}{\int }_{-1}^{+1}dx=\frac{{\mu }_{0}nI{r}^2\left(2l\right)}{2a^3}$

The magnetic moment of solenoid $m\left(NIA\right)=\left(n2I\right)I.\pi r^2$
$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2m}{a^3}$

This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along axial line is similar to that of a bar magnet for far off axial points.